The heat and wave equations in 3D
18.303 Linear Partial Differential Equations
Matthew J. Hancock
Fall 2004
1 3DHeatEquation
[Oct 27, 2004]
Ref: §1.5 Haberman
Consider an arbitrary 3D subregion V of R3 (V ⊆ R3), with temperature u (x, t)
defined at all points x = (x, y, z)
V . We generalize the ideas of 1-D heat flux to
find an equation governing u. The heat energy in the subregion is defined as
heat energy＝
Recall that conservation of energy implies
rate of change
heat energy into V from
heat energy generated
of heat energy
boundaries per unit time
in solid per unit time
We desire the heat flux through the boundary S of the subregion V, which is
the normal component of the heat flux vector φ, φ ·û, where û is the outward unit
normal at the boundary S. Hats on vectors denote a unit vector, | û | = 1 (length 1).
If the heat flux vector φ is directed inward, then φ ·û < 0 and the outward flow of
heat is negative. To compute the total heat energy flowing across the boundaries,
we sum φ ·û over the entire closed surface S, denoted by a double integral
Therefore, the conservation of energy principle becomes
1
1.1 Divergence Theorem (a.k.a. Gauss’s Theorem)
For any volume V with closed smooth surface S,
where A is any function that is smooth (i.e. continuously differentiable) for x
Note that
V.
where êx, êy, êz are the unit vectors in the x, y, z directions, respectively. The
divergence of a vector valued function F = (Fx, Fy, Fz) is
The Laplacian of a scalar function F is
Applying the Divergence theorem to (1) gives
Since V is independent of time, the integrals can be combined as
Since V is an arbitrary subregion of R3 and the integrand is assumed continuous, the
integrand must be everywhere zero,
1.2 Fourier’s Law of Heat Conduction
The 3D generalization of Fourier’s Law of Heat Conduction is (see Appendix to 1.5,
pp 32, Haberman)
where K0 is called the thermal diffusivity. Substituting (3) into (2) gives
2
where κ = K0/ (cρ). This is the 3D Heat Equation. Normalizing as for the 1D case,
the dimensional Heat Equation (4) becomes (dropping tildes)
where q = l2Q/ (κcρ) = l2Q/K0.
2 3DWaveequation
The 1D wave equation can be generalized (Haberman, §4.5) to a 3D wave equation,
in scaled coordinates,
3 Separation of variables in 3D
[Oct 29, 2004]
Ref: Haberman, Ch 7.
We consider simple subregions D ⊆ R3. We assume the boundary conditions are
zero, u = 0 on ӘD, where ӘD denotes the closed surface of D (assumed smooth).
The 3D Heat Problem is
The 3D wave problem is
We separate variables as
3
The 3D Heat Equation implies
where λ = const since the l.h.s. depends solely on t and the middle X__/X depends
solely on x. The 3D wave equation becomes
On the boundaries,
3.1 Sturm-Liouville problem
Both the 3D Heat Equation and the 3D Wave Equation lead to the Sturm-Liouville
problem
3.2 Positive, real eigenvalues (for Type I BCs)
Ref: §7.4 Haberman
For the Type I BCs assumed here (u (x, t) = 0, for x
ӘD), we now show that
all eigenvalues are positive. To do so, we need a result that combines some vector
calculus with the Divergence Theorem. From vector calculus, for any scalar function
G and vector valued function F,
Using the divergence theorem,
Substituting (13) into (14) gives
Choosing G = v and F = ▽v, for some function v, we have
4
Result (16) holds for any smooth function v defined on a volume V with closed smooth
surface S.
We now apply result (16) to a solution X (x) of the Sturm-Liouville problem (12).
Letting v = X (x), S = ӘD and V = D, Eq. (16) becomes
Since X (x) = 0 for x
ӘD,
Also, from the PDE in (12),
Substituting (18) and (19) into (17) gives
For non-trivial solutions, X≠0 at some points in D and hence by continuity of X,
Thus (20) can be rearranged,
Since X is real, the the eigenvalue λ is also real.
If ▽X = 0 for all points in D, then integrating and imposing the BC X (x) = 0
for x
ӘD gives X = 0 for all x
D, i.e. the trivial solution. Thus ▽X is nonzero
at some points in D, and hence by continuity of ▽X,
Thus,
from (21), λ > 0.
4 Solution for T (t)
Suppose that the Sturm-Liouville problem (12) has eigen-solution Xn (x) and eigenvalue
λn, where Xn (x) is non-trivial. Then for the 3D Heat Problem, the problem
for T (t) is, from (10),
with solution
5
6
and the corresponding solution to the PDE and BCs is
For the 3D Wave Problem, the problem for T (t) is, from (11),
with solution
and the corresponding normal mode is (x, t) = Xn (x) Tn (t).
5 Uniqueness of the 3D Heat Problem
We now prove that the solution of the 3D Heat Problem
is unique. Let u1, u2 be two solutions. Define v = u1 － u2. Then v satisfies
Let
V (t)
0 since the integrand v2 (x, t)
0 for all (x, t). Differentiating in time gives
Substituting for vt from the PDE yields
By result (16),
7
But on ӘD, v = 0, so that the first integral on the r.h.s. vanishes. Thus
Also, at t = 0,
Thus V (0) = 0, V (t)
0 and dV/dt
0, i.e. V (t) is a non-negative, non-increasing
function that starts at zero. Thus V (t) must be zero for all time t, so that v (x, t)
must be identically zero throughout the volume D for all time, implying the two
solutions are the same, u1 = u2. Thus the solution to the 3D heat problem is unique.
6 Orthogonality of eigen-solutions to
Sturm-Liouville problem
Ref: §7.4 Haberman
Suppose v1, v2 are two eigen-functions with eigenvalues λ1, λ2 of the 3D SturmLiouville problem
Result (15) applied to G = v1 and F = ▽2v2 gives
Since v1 = 0 on ӘD and ▽2v2 = λ2v2, we have
Similarly, applying result (15) to G = v2 and F = ▽v1 gives
Subtracting (27) from (26) gives
Thus if λ1 ≠ λ2,
and the eigen-functions v1, v2 are orthogonal.
8
9
7 Heat and Wave problems on a 2D rectangle
[Nov. 1, 2004]
Ref: §7.3 Haberman
7.1 Sturm-Liouville Problem on a 2D rectangle
We now consider the special case where the subregion D is a rectangle
The Sturm-Liouville Problem (12) becomes
Note that in the PDE (29), λ is positive a constant (we showed above that λ had to
be both constant and positive). We employ separation of variables again, this time
in x and y: substituting v (x, y) = X (x) Y (y) into the PDE (29) and dividing by
X (x) Y (y) gives
Since the l.h.s. depends only on y and the r.h.s. only depends on x, both sides must
equal a constant, say μ,
The BCs (30) and (31) imply
To have a non-trivial solution, Y (y) must be nonzero for some y
must be nonzero for some x
must have
[0, y0] and X (x)
[0, x0], so that to satisfy the previous 2 equations, we
The problem for X (x) is the 1D Sturm-Liouville problem
10
We solved this problem in the chapter on the 1D Heat Equation. We found that for
non-trivial solutions, μ had to be positive and the solution is
The problem for Y (y) is
where ν = λ － μ. The solutions are the same as those for (34), with ν replacing μ:
The eigen-solution of the 2D Sturm Liouville problem (29) – (31) is
with eigenvalue
See plots in Figure 7.3.2 (p 284) Haberman of vmn (x, y) for various m, n.
7.2 Solution to heat equation on 2D rectangle
The heat problem on the 2D rectangle is the special case of (7),
where D is the rectangle D = {(x, y) : 0
x
x0, 0
y
y0}. We reverse the
separation of variables (9) and substitute solutions (23) and (38) to the T (t) problem
(22) and the Sturm Liouville problem (29) – (31), respectively, to obtain
11
To satisfy the initial condition, we sum over all m, n to obtain the solution, in
general form,
Setting t = 0 and imposing the initial condition u (x, y, 0) = f (x, y) gives
Where
are the eigen-functions of the 2D Sturm Li-ouville
problem on a rectangle, (29) – (31). Multiplying both sides by v ˆmˆn (x, y) (ˆm,n = 1, 2, 3, ...)
and integrating over the rectangle D gives
where dA = dxdy. Note that
Thus (39) becomes
Since m,n are dummy variables, we replace ˆm by m and ˆn by n, and rearrange to
obtain
7.3 Solution to wave equation on 2D rectangle
Application: waves on a 2D rectangular membrane (§7.3 Haberman)
12
The solution to the wave equation on the 2D rectangle follows similarly. The
general 3D wave problem (8) becomes
where D is the rectangle D = {(x, y) : 0
x
x0, 0
y
y0}. We reverse the
separation of variables (9) and substitute solutions (25) and (38) to the T (t) problem
(24) and the Sturm Liouville problem (29) – (31), respectively, to obtain
To satisfy the initial condition, we sum over all m, n to obtain the solution, in
general form,
Setting t = 0 and imposing the initial conditions
gives
where
are the eigen-functions of the 2D Sturm Liouville
problem on a rectangle, (29) – (31). As above, multiplying both sides by
13
v ˆmˆn (x, y) (m, n = 1, 2, 3, ...) and integrating over the rectangle D gives
8 Heat and Wave equations on a 2D circle
[Nov 3, 2004]
Ref: §7.7 Haberman
We now consider the special case where the subregion D is the unit circle (we may
assume the circle has radius 1 by choosing the length scale l for the spatial coordinates
as the original radius):
The Sturm-Liouville Problem (12) becomes
where we already know λ is positive and real. It is natural to introduce polar coordinates
via the transformation
for
You can verify that
The PDE becomes
The BC (42) requires
We use separation of variables by substituting
14
into the PDE (43) and multiplying by r2/ (R(r)H (θ)) and then rearranging to obtain
Again, since the l.h.s. depends only on r and the r.h.s. on θ, both must be equal to
a constant μ,
The BC (44) becomes
which, in order to obtain non-trivial solutions (H (θ) _= 0 for some θ), implies
In the original (x, y) coordinates, it is assumed that v (x, y) is smooth (i.e. continuously
differentiable) over the circle. When we change to polar coordinates, we need
to introduce an extra condition to guarantee the smoothness of v (x, y), namely, that
Substituting (45) gives
The solution v (x, y) is also bounded on the circle, which implies R(r) must be
bounded for 0
r
1.
The problem for H (θ) is
You can show that for μ < 0, we only get the trivial solution H (θ) = 0. For μ = 0,
we have H (θ) = const, which works. For μ > 0, non-trivial solutions are found only
when μ = m2,
Thus, in general, we may assume λ = m2, for m = 0, 1, 2, 3, ...
The equation for R(r) in (46) becomes
15
Rearranging gives
We know already that λ > 0, so we can let
so that (51) becomes
The ODE is called Bessel’s Equation which, for each m = 0, 1, 2, ... has two linearly
independent solutions, Jm (s) and Ym (s), called the Bessel functions of the first and
second kinds, respectively, of order m. The function Jm (s) is bounded at s = 0;
the function Ym (s) is unbounded at s = 0. The general solution to the ODE is
where cmn are constants of integration. Our boundedness
criterion
implies C2m=0 Thus
The Bessel Function Jm (s) of the first kind of order m has power series
Jm (s) can be expressed in many ways, see Handbook of Mathematical
Functions by Abramowitz and Stegun, for tables, plots, and equations. The fact
that Jm (s) is expressed as a power series is not a drawback. It is like sin (x) and
cos (x), which are also associated with power series. Note that the power series (53)
converges absolutely for all s
0 and converges uniformly on any closed set s
L]. To see this, note that each term in the sum satisfies
Note that the sum of numbers
converges by the Ratio Test, since the ratio of successive terms in the sum is
16
[0,
Thus for k >N = [L/2],
Since the upper bound is less than one and is independent of the summation index k,
then by the Ratio test, the sum converges absolutely. By the Weirstrass M-Test, the
infinite sum in (53) converges uniformly on [0, L]. Since L is arbitrary, the infinite
sum in (53) converges uniformly on any closed subinterval [0, L] of the real axis.
Each Bessel function Jm (s) has an infinite number of zeros (roots) for s > 0. Let
Jm,n be the n’th such zero for the function Jm (s). Note that
The second BC requires
This has an infinite number of solutions, namely
the eigenvalues are
= Jm,n for n = 1, 2, 3, .... Thus
with corresponding eigen-functions Jm (rJm,n). The separable solutions are thus
8.1 Solution to heat equation on the 2D circle
The heat problem on the 2D circle is the special case of (7),
where D is the circle D = {(x, y) : x2 + y2
1}. We reverse the separation of variables
(9) and substitute solutions (23) and (38) to the T (t) problem (22) and the
Sturm Liouville problem (41) – (42), respectively, to obtain
17
where vmn is given in (54).
To satisfy the initial condition, we sum over all m, n to obtain the solution, in
general form,
Setting t = 0 and imposing the initial condition u (x, y, 0) = f (x, y) gives
We can use orthogonality relations to find Amn.
9
The Heat Problem on a square with inhomogeneous BC
[Nov 8, 2004]
We now consider the case of the heat problem on a 2D square of scaled side length
1, where a hot spot exists on the left side:
where the hot spot is confined to the left side: 0
y0
ε/2
y
y0 +ε/2
1.
As in the 1D case, we first find the equilibrium solution uE (x, y), which satisfies the
PDE and the BCs,
We proceed via separation of variables: uE (x, y) = X (x) Y (y), so that the PDE
becomes
where λ is constant since the l.h.s. depends only on x and the middle only on y. The
BCs are
18
and
We first solve for Y (y), since we have 2 easy BCs:
The non-trivial solutions, as we have found before, are Yn = sin(nπy) with λn = n2π2,
for each n = 1, 2, 3, ... Now we consider at X (x):
and hence
An equivalent and more convenient way to write this is
Imposing the BC at x = 1 gives
and hence
Thus the equilibrium solution to this point is
You can check that this satisfies the BCs on x = 1 and y = 0, 1. Also, from the BC
on x = 0, we have
Multiplying both sides by sin (mπy) an integrating in y gives
From the orthogonality of sin’s, we have
19
Thus,
Thus
To solve the transient problem, we proceed as in 1-D by defining the function
so that v (x, y, t) satisfies
9.1 First term approximation
To approximate the equilibrium solution uE (x, y), note that
For sufficiently large n, we have
Thus the terms decrease in magnitude (x > 0) and hence uE (x, y) can be approximated
the first term in the series,
A plot of sinh (π (1 － x)) sin (πy) is given below. The temperature in the center of
the square is approximately
20
Figure 1: Plot of sinh (π(1-x))sin (πy).
9.2 Easy way to find steady-state temperature at center
For y0 = 1/2 and ε = 1, we have
It turns out there is a much easier way to derive this last result. Consider a plate
with BCs u = u0 on one side, and u = 0 on the other 3 sides. Let
Rotating the plate by 90o will not alter
since this is the center of the plate.
Let uEsum be the sums of the solutions corresponding to the BC u = u0 on each of the
four different sides. Then by linearity, uEsum = u0 on all sides and hence uEsum = u0
across the plate. Thus
21
9.3 Placement of hot spot for hottest steady-state center
Note that
for small ε. Thus the steady-state center temperature is hottest when the hot spot is
placed in the center of the side, i.e. y0 = 1/2.
10 Heat problem on a circle with inhomogeneous
BC
[Nov 10, 2004]
Consider the heat problem
where D = {(x, y) : x2 + y2
1} is a circle (disc) of radius 1. To solve the problem,
we must first introduce the steady-state u = uE (x, y) which satisfies the PDE and
BCs,
As before, switch to polar coordinates via
for
22
The problem for uE becomes
where ˆg (θ) = g (x, y) for (x, y)
We separate variables
ӘD and θ = arctan (y/x).
and the PDE becomes
Since the l.h.s. depends only on r and the r.h.s. on θ, both must be equal to a
constant μ,
The problem for H (θ) is, as before,
with eigen-solutions
Hm (θ) = am cos (mθ) + bm sin (mθ), m= 0, 1, 2, ...
The problem for R(r) is
Try R(r) = rα to obtain the auxiliary equation
α (α － 1) + α － m = 0
2
whose solutions are α = ±m. Thus for each m, the solution is
Our boundedness criterion (57) implies c2 = 0.
Hence
23
where cm are constants to be found by imposing the BC (58). The separable solutions
satisfying the PDE (55) and conditions (56) to (57), are
The full solution is the infinite sum of these over m,
We still need to find the Am, Bm.
Imposing the BC (58) gives
The orthogonality relations are
Multiplying (60) by sin nθ or cos nθ nd applying these orthogonality relations gives
10.1 Hot spot on boundary
Suppose
which models a hot spot on the boundary. The Fourier coefficients in (61) are thus
Thus the steady-state solution is
24
Figure2: Setup for hot spot problem on circle.
10.2 Interpretation
[Nov 12, 2004]
The convergence of the infinite series is rapid if r << 1. If r
1, many terms are
required for accuracy.
The center temperature (r = 0) at equilibrium (steady-state) is
i.e., the mean temperature of the circumference. This is a special case of the Mean
Value Property of solutions to Laplace’s Equation ▽2u = 0.
We now consider plots of uE (x, y) for some interesting cases. We draw the level
curves (isotherms) uE = const as solid lines. Recall from vector calculus that the
gradient of uE, denoted by ▽uE, is perpendicular to the level curves. Recall also
from the physics that the flux of heat is proportional to ▽uE. Thus heat flows along
the lines parallel to ▽uE. Note that the heat flows even though the temperature is in
steady-state. It is just that the temperature itself at any given point does not change.
We call these lines the “heat flow lines” or the “orthogonal trajectories”, and draw
these as dashed lines in the figure below.
Note that lines of symmetry correspond to (heat) flow lines. To see this, let nl
be the normal to a line of symmetry. Then the flux at a point on the line is ▽u · nl.
Rotate the image about the line of symmetry. The arrow for the normal to the line of
25
Figure 3: Plots of steady-state temperature due to a hot segment on boundary.
symmetry is now pointing in the opposite direction, i.e. －nl, and the flux is －▽u·nl.
But since the solution is the same, the flux across the line must still be ▽u · nl. Thus
which implies ▽u·nl = 0. Thus there is no flux across lines of symmetry. Equivalently,
▽u is perpendicular to the normal to the lines of symmetry, and hence ▽u is parallel
to the lines of symmetry. Thus the lines of symmetry are flow lines. Identifying the
lines of symmetry help draw the level curves, which are perpendicular to the flow
lines. Also, lines of symmetry can be thought of as an insulating boundary, since
▽u · nl = 0. See Problem 2 of Assnt 5.
( i ) θ0 = 0. Then
Use the BCs for the boundary. Note that the solution is symmetric with respect
to the y-axis (i.e. even in x). The solution is discontinuous at {y = 0, x = ±1}, or
{r = 1, θ = 0, π}. See plot.
( ii ) －π < θ0 < －π/2. The sum for uE is messy, so we use intuition. We start
with the boundary conditions and use continuity in the interior of the plate to obtain
a qualitative idea of the level curves and heat flow lines. See plot.
(iii) θ0 → －π+ (a heat spot). Again, use intuition to obtain a qualitative sketch
of the level curves and heat flow lines. Note that the temperature at the hot point is
infinite. See plot.
26
11 Mean Value Property
Theorem [Mean Value Property] Suppose v (x, y) satisfies Laplace’s equation in
a 2D domain D,
Then at any point (x0, y0) in D, v equals the mean value of the temperature around
any circle centered at (x0, y0) and contained in D,
Note that the curve {(x0 + Rcos θ, y0 + Rsin θ) : －π
radius R centered at (x0, y0).
θ < π} traces the circle of
Proof: To prove the Mean Value Property, we first consider Laplace’s equation
(62) on the unit circle centered at the origin (x, y) = (0, 0). We already solved this
problem, above, when we solved for the steady-state temperature uE that took the
value g (θ) on the boundary. The solution is Eqs. (59) and (61). Setting r = 0 in
(59) gives the center value
On the boundary of the circle (of radius 1), (x, y) = (cosθ, sin θ) and the BC implies
that uE = g (θ) on that boundary. Thus, g (θ) = uE (cos θ, sin θ) and (64) becomes
To prove the Mean Value Property, we consider the region
which is a circle of radius R centered at (x, y) = (x0, y0). Since B(x0,y0) (R) ⊆ D,
Laplace’s equation (62) holds on this circle. Thus
We make the change of variable
to map the circle B(x0,y0) (R) into the unit circle {(x,y) : x2 +y2
equation (66) becomes
27
1}. Laplace’s
where
The solution is given by Eqs. (59) and (61), and we
found the center value above in Eq. (65). Reversing the change of variable (67) in
Eq. (65) gives
as required. ■
For the heat equation, the Mean Value Property implies the equilibrium temperature
at any point (x0, y0) in D equals the mean value of the temperature around any
circle centered at (x0, y0) and contained in D.
12 Maximum Principle
Theorem [Maximum Principle] Suppose v (x, y) satisfies Laplace’s equation in a
2D domain D,
Then the function v takes its maximum and minimum on the boundary of D, ӘD.
Proof: Let (x0, y0) be any interior point of D, i.e. (x, y) is not on the boundary
ӘD of D. The Mean Value Property implies that for any circle of radius R centered
at (x0, y0),
But the the average of a set of numbers is always between the minimum and maximum
of those numbers. Thus the average value of v (x, y) on the boundary must be between
the minimum and maximum value of v (x, y) on the boundary, and hence v (x0, y0) is
between the minimum and maximum values of v (x, y) on the boundary. ■
For the heat equation, this implies the equilibrium temperature cannot attain its
maximum in the interior, unless temperature is constant everywhere.
28
13 Eigenvalues on different domains
[Nov 15, 2004]
Definition Rayleigh Quotient
Theorem Given a domain D ⊆ R3 and any function v that is piecewise smooth
on D, non-zero at some points on the interior of D, and zero on all of ӘD, then the
smallest eigenvalue of the Laplacian on D satisfies
and R(h) = λ if and only if h (x) is an eigen-solution of the Sturm Liouville problem
on D.
Sketch Proof: We use result (16) derived for any smooth function v defined on
a volume V with closed smooth surface S.
In the statement of the theorem, we assumed that v = 0 on ∂D, and hence
Let {φn} be an orthonormal basis of eigen-functions on D, i.e. all the functions φn
which satisfy
and
We can expand v in the eigen-functions,
where the An are constants. Assuming we can differentiate the sum termwise, we
have
29
The orthonormality property (i.e., the orthogonality property with
implies
and, from (70),
Substituting (72) into (69) gives
Substituting (71) and (73) into (68) gives
We assume the eigen-functions are arranged in increasing order. In particular, λn
λ1. Thus
Also, equality holds (i.e., R(v) = λ1) if and only if v is an eigen-function for the
eigenvalue λ1. Otherwise there will be a λn > λ1 such that An _= 0 and R(v) > λ1.
The above is not a complete proof, because we have not shown that the sums converge
or that they can be differentiated termwise. ■
Theorem If two domains
and D in R2 satisfy
In other words, the domain
that contains the sub-domain D is associated with a
smaller eigenvalue.
Proof: Note that the Sturm-Liouville problems are
30
Let v1 be the eigen-function corresponding to λ1 on D. Then, as we have proven
before,
where R(v) is the Rayleigh Quotient. Extend the function v1 continuously from D
to
to obtain a function v1 on
which satisfies
The extension is continuous, since v1 is zero on the boundary of D. Applying the
previous theorem to the region
and function v1 (which satisfies all the requirements
of the theorem) gives
Equality happens only if v1 is the eigen-function corresponding toλ1.
Useful fact [stated without proof]: the eigen-function(s) corresponding to the
smallest eigenvalueλ1 on
are nonzero in the interior of
. This is the crux of the
proof. Haberman does not prove this in general, but states it on p. 164.
From the useful fact, v1 cannot be an eigen-function corresponding toλ1 on
,
since it is zero in the interior of
(outside D). Thus, as the previous theorem states,
Since v1 = 0 outside D, the integrals over
integrals over D, where v1 = v1, and hence
in the Rayleigh quotient reduce to
Combining (74), (75), and (76) gives the result,
■
Example: Consider two regions, D1 is a rectangle of length x0, height y0 and D2
is a circle of radius R. Recall that the smallest eigenvalue on the rectangle D1 is
31
The smallest eigenvalue on the circle of radius 1 is
where the first zero
of the Bessel function J0 (s) of the first kind is J0,1 = 2.4048. Since
multiplied
r in the Bessel function, then for a circle of radius R, we’d rescale by the change of
variable r = r/R, so that
where r goes from 0 to 1. Thus
on the circle of radius R, the smallest eigenvalue is
Suppose the rectangle is actually a square of side length 2R. Then
Thus, λ11 < λ01, which confirms the second theorem, since D2 ⊂ D1, i.e., the circle
is contained inside the square.
Now consider the function
You can show that
and
This confirms the first theorem, since v (r) is smooth on D2, v (R) = 0 (zero on the
boundary of D2), and v is nonzero in the interior.
13.1 Faber-Kahn inequality
Thinking about the heat problem on a 2D plate, what shape of plate will cool the
slowest? It is a geometrical fact that of all shapes of equal area, the circle (disc) has
the smallest circumference. Thus, on physical grounds, we expect the circle to cool
the slowest. Faber and Krahn proved this in the 1920s.
Faber-Kahn inequality For all domains D ⊂ R2 of equal area, the disc has the
smallest first eigenvalue λ1.
Example. Consider the circle of radius 1 and the square of side length
. Then
both the square and circle have the same area. The first eigenvalues for the square
and circle are, respectively,
and hence λ1SQ > λ1CIRC, as the Faber-Kahn inequality states.
32
33
14 Nodal lines
Consider the Sturm-Liouville problem
Nodal lines are the curves where the eigen-functions of the Sturm-Liouville problem
are zero. For the solution to the vibrating membrane problem, the normal modes
unm (x, y, t) are zero on the nodal lines, for all time. These are like nodes on the 1D
string. Here we consider the nodal lines for the square and the disc (circle).
14.1 Nodal lines for the square
[Nov 17, 2004]
See Haberman, p. 292.
For the square, the eigen-functions and eigenvalues are a special case of those we
found for the rectangle, with side length x0 = y0 = a,
The nodal lines are the lines on which vmn (x, y) = 0, and are
for m, n
2. Note that v11 (x, y) has no nodal lines on the interior - it is only zero
on boundary ӘD. Since λmn = λnm, then the function fnm = Avmn +Bvnm is also an
eigen-function with eigenvalue λmn, for any constants A, B. The nodal lines for fnm
can be quite interesting.
Examples: we draw the nodal lines on the interior and also the lines around the
boundary, where vnm = 0.
(i) m = 1, n = 1.
This is positive on the interior and zero on the boundary. Thus the nodal lines are
simply the square boundary ӘD
(ii) m = 1, n = 2.
The nodal lines are the boundary ӘD and the horizontal line y = a/2.
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(iii) m = 1, n = 3
The nodal lines are the boundary Ә D and the horizontal lines y = a/3, 2a/3.
(iv) m = 3, n = 1
The nodal lines are the boundary Ә D and the vertical lines x = a/3, 2a/3.
(v) Consider v13 － v31. Since λ31 = λ13 = 10π2/a2, this is a solution to
To find the nodal lines, we use the identity sin 3θ = (sinθ) (3 － 4 sin2 θ) to write
The nodal lines are the boundary of the square, ӘD, and lines such that
i.e.,
Note that
if
for any integer k. Thus the nodal lines are given by
Hence the nodal lines are
For all intergers l. We’re only concerned with the nodal lines that intersect the interior of
the square plate:
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Thus the nodal lines of v13 － v31 are the sides and diagonals of the square
Let DT be the isosceles right triangle whose hypotenuse lies on the bottom horizontal
side of the square The function v13 － v31 is zero on the boundary ӘDT , positive on the
interior of DT , and thus satisfies the Sturm-Liouville problem
Hence v13 － v31 is the eigen-function corresponding to the first eigenvalue λ13 of
the Sturm-Liouville problem on the triangle DT . In this case, we found the eigenfunction
without using separation of variables, which would have been complicated on the triangle.
NOTE: this is half the triangle considered in problem 4 of Assnt 5, but you use a similar
solution method.
(vi) With v13, v31 given above, adding gives
The nodal lines for v13 + v31 are thus the square boundary and the closed nodal line
defined by
Let Dc be the area contained within this closed nodal line. The function －(v13 + v31)
is zero on the boundary ӘDc, positive on the interior of Dc, and thus satisfies the
Sturm-Liouville problem
Hence －(v13 + v31) is the eigen-function corresponding to the first eigenvalue λ13 of
the Sturm-Liouville problem on Dc.
(vii) Find the first eigenvalue on the right triangle
Note that separation of variables is ugly, because you’d have to impose the BC
We proceed by placing the triangle inside a rectangle of horizontal and vertical side
lengths 1 and √3, respectively. The sides of the rectangle coincide with the perpendicular
sides of the triangle. Thus, all eigen-functions vmn for the rectangle are
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already zero on two sides of the triangle. However, any particular eigen-function vmn
will not be zero on the triangle’s hypotenuse, since all the nodal lines of vmn are
horizontal or vertical. Thus we need to add multiple eigen-functions. However, to
satisfy the Sturm-Liouville problem, all the eigen-functions must be associated with
the same eigenvalue. For the rectangle with side lengths √3 and 1, the eigenvalues
are given by
We create a table of 3m2 + n2 and look for eigenvalues that have multiple eigenfunctions.
The smallest value of 3m2+n2 that is the same for multiple sets of (m, n) is 28. Thus
has multiple eigen-functions. We add the corresponding eigen-functions and set them
to zero along y = √3x,
If we can find the constants A, B, C then we’re done - we’ve found the first (smallest)
eigenvalue 28π2/3 and eigen-function Av31+Bv24+Cv15 on the triangle D. Otherwise,
if we can’t solve for A, B, C, we look for the next largest value of 3m2 + n2 that
repeats, and try again.
14.2 Nodal lines for the disc (circle)
[See Haberman, p. 321]
For the disc of radius 1, we found the eigen-functions and eigenvalues to be
vmnS = Jm (rJm,n) sin mθ, vmnC = Jm (rJm,n) cos mθ
with
The nodal lines are the lines on which vmnC = 0 or vmnS = 0.
Examples.
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(i) m = 0, n = 1.
The nodal lines are the boundary of the disc (circle of radius 1).
(ii) m = 0, n = 2.
The nodal lines are two concentric circles, one of radius r = 1, the other or radius
r = J0,1/J0,2 < 1.
(iii) m = 1, n = 1 and sine.
The nodal lines are the boundary (circle of radius 1) and the line θ = －π, 0, π
(horizontal diameter)
15 Steady-state temperature in a 3D cylinder
Suppose a 3D cylinder of radius a and height L has temperature u (r, θ, z, t). We
assume the axis of the cylinder is on the z-axis and (r, θ, z) are cylindrical coordinates.
Initially, the temperature is u (r, θ, z, 0). The ends are kept at a temperature of u = 0
and sides kept at u (a, θ, z, t) = g (θ, z). The steady-state temperature uE (r, θ, z) in
the 3D cylinder is given by
In cylindrical coordinates (r, θ, z), the Laplacian operator becomes
We separate variables as
so that (77) becomes
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Rearranging gives
where λ is constant since the l.h.s. depends only on r, θ while the middle depends
only on z.
The function Z (z) satisfies
The BCs at z = 0, L imply
To obtain non-trivial solutions, we must have
As we’ve shown many times before, the solution for Z (z) is, up to a multiplicative
constant,
And this also shows that the constant λ = λn.
Multiplying Eq. (79) by r2 gives
where μ is constant since the l.h.s. depends only on r and the middle only on θ.
Note: solving for Hn (θ) looks easier, until you realize that we don’t have nice BCs
on Hn (θ).
Multiplying (80) by Rn (r) gives
Change variables to s =
, Rn (s) = Rn (r), so that
If μ = m2 for some m = 1, 2, 3, ..., we have
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which is the Modified Bessel Equation with tabulated solutions Im (s) and Km (s)
called the modified Bessel functions of order m of the first and second kinds, respectively
We assume μ = m2, so that
Transforming back to r =
gives
Since the Im’s are regular (bounded) at r = 0, while the Km’s are singular (blow up),
and since Rmn (r) must be bounded, we must have c2m = 0, or
The corresponding solutions for H (θ) satisfy
and hence Hm (θ) = Am cos mθ + Bm sin mθ.
Thus the general solution is, by combining constants,
In theory, we can now impose the condition u (a, θ, z) = g (θ, z) and find Amn, Bmn
using orthogonality of sin, cos.
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